There is one big-daddy among the trig identities that you need to learn right now, if you don't know it already:

$$\sin^2(x) + \cos^2(x) = 1$$

This is the identity that nearly all of the others spring from.

There are some more definitions: $tan(x) = \frac{\sin(x)}{\cos(x)}$, which is one of the first things you learn; there are also three minor trig functions, namely:

- $\sec(x) = \frac{1}{\cos(x)}$
- $\cosec(x) = \frac{1}{\sin(x)}$ [This is sometimes written as "$\csc$".]
- $\cot(x) = \frac{1}{\tan(x)}$

### Dividing through trig identities...

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From there, you can think about what happens if you divide the big daddy by either $\cos^2(x)$ or $\sin^2(x)$.

$$\frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}$$

$$\tan^2(x) + 1 = \sec^2(x)$$

$$\frac{\sin^2(x)}{\sin^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} = \frac{1}{\sin^2(x)}$$

$$1 + \cot^2(x) = \cos^2(x)$$

I don't know these two trig identities by heart, although I probably ought to: I work them out on the fly, since they don't take very long.

Why do you care about such esoteric trig identities? Well, aside from them being the way to solve many of the trig problems in C2 and C3, they're also useful for integrating. You don't have $\tan^2(x)$ or $\cot^2(x)$ in your formula books - but you do have $\sec^2(x)$ and $\cosec^2(x)$. That means, if you have to integrate $\tan^2(x)$ or $\cot%2(x)$, you can just rearrange along these lines:

$\tan^2(x) = \sec^2(x) - 1$, so

$\int \tan^2(x) dx = \int (\sec^2(x) - 1) dx = \tan(x) - x + c$

So that's accounted for all of the squared minor trig functions and $\tan^2(x)$ to boot - but how about $\sin^2(x)$ and $\cos^2(x)$?

... well, I'll leave that for another post. There's more than one way to do it.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.